Here you will learn what is the eccentricity of ellipse formula and how to find eccentricity with examples.
Letโs begin โ
Eccentricity of Ellipse Formula
(i) For the ellipse \(x^2\over a^2\) + \(y^2\over b^2\) = 1, a > b
we have,ย \(b^2\) = \(a^2(1 โ e^2)\)
\(\implies\)ย \(e^2\) = 1 โ \(b^2\over a^2\)
\(\implies\)ย eccentricity (e) = \(\sqrt{1 โ {b^2\over a^2}}\)
(ii) For the ellipse \(x^2\over a^2\) + \(y^2\over b^2\) = 1, a < b
eccentricity (e) = \(\sqrt{1 โ {a^2\over b^2}}\)
Also Read : Different Types of Ellipse Equations and Graph
Example : For the given ellipses, find the eccentricity.
(i)ย \(16x^2 + 25y^2\) = 400
(ii)ย \(x^2 + 4y^2 โ 2x\) = 0
Solution :
(i)ย We have,
\(16x^2 + 25y^2\) = 400 \(\implies\) \(x^2\over 25\) + \(y^2\over 16\),
where \(a^2\) = 25 and \(b^2\) = 16 i.e. a = 5 and b = 4
Clearly a > b,
Therefore, the eccentricity of ellipse (e) = \(\sqrt{1 โ {b^2\over a^2}}\)
e = \(\sqrt{1 โ 16/25}\) = \(3\over 5\)
(ii) We have,
\(x^2 + 4y^2 โ 2x\) = 0
\(\implies\) \((x โ 1)^2\) + 4\((y โ 0)^2\) = 1
\(\implies\)ย \((x โ 1)^2\over 1^2\) + \((y โ 0)^2\over (1/2)^2\) = 1
Here, a = 1 and b = 1/2
Clearly a > b,
Therefore, the eccentricity of ellipse (e) = \(\sqrt{1 โ {b^2\over a^2}}\)
e = \(\sqrt{1 โ 1/4}\) = \(\sqrt{3}\over 2\)
